Saturday 14 July 2018

trigonometry - Proof of $sin^2 x+cos^2 x=1$ using Euler's Formula



How would you prove $\sin^2x + \cos^2x = 1$ using Euler's formula?




$$e^{ix} = \cos(x) + i\sin(x)$$



This is what I have so far:



$$\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$$



$$\cos(x) = \frac{1}{2} (e^{ix}+e^{-ix})$$


Answer



Multiply $\mathrm e^{\mathrm ix}=\cos(x)+\mathrm i\sin(x)$ by the conjugate identity $\overline{\mathrm e^{\mathrm ix}}=\cos(x)-\mathrm i\sin(x)$ and use that $\overline{\mathrm e^{\mathrm ix}}=\mathrm e^{-\mathrm ix}$ hence $\mathrm e^{\mathrm ix}\cdot\overline{\mathrm e^{\mathrm ix}}=\mathrm e^{\mathrm ix-\mathrm ix}=1$.



No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...